Sunday, 2 June 2019

quantum field theory - Sign ambiguity when going from position to momentum space evaluating Feynman diagrams


When calculating a simple diagram I came across an ambiguity in the conservation of momentum, i.e. it seems to me that the particle could come out of the process with opposite momentum with respect to its initial state. This might be a complete triviality, in that case I apologize, but I can't find my mistake. Take the following Feynman diagram in λϕ4 theoryenter image description here


Call x and y the two external points, z1 and z2 the two internal points. Using Feynman's rules in position space this is proportional to the integral


dz1dz2DF(xz1)DF(z1z2)2DF(z2z2)DF(z1y)


Where DF is the Feynman propagator of the Klein Gordon field


DF(xy)=d4p(2π4)ip2m2+iεeip(xy)



Inserting this into the diagram, and writing dq=d4px(2π)4d4py(2π)4d4p1(2π)4d4p2(2π)4d4p3(2π)4 for the sake of brevity


dz1dz2DF(xz1)DF(z1z2)2DF(z2z2)DF(z2y)==dz1dz2dqeipx(xz1)p2xm2+iεeip1(z1z2)p21m2+iεeip2(z1z2)p22m2+iε1p23m2+iεeipy(z1y)p2ym2+iε==dz1dz2dqeiz1(px+p1+p2+py)p2xm2+iεeiz2(p1p2)p21m2+iεei(pxxpyy)p22m2+iε1p23m2+iε1p2ym2+iε


The momenta in the exponentials are gonna give δ(4)(pxp1p2py)δ(4)(p1+p2)


Which implies conservation of momentum, and everyone is happy. But let's get back to the first expression of the integral, where I wrote


dz1dz2DF(xz1)DF(z1z2)2DF(z2z2)DF(z1y)


I could have written


dz1dz2DF(xz1)DF(z1z2)2DF(z2z2)DF(yz1)


because the propagator is symmetric. This seemingly harmless change changes the exponential that yields the delta in


eiz1(px+p1+p2py)


and the conservation of momentum will be δ(4)(pxp1p2+py)δ(4)(p1+p2)



which in the end implies px=py, or, the particle comes out with opposite momentum, which seems weird to me. Have I made a mistake or is the sign of the momentum not important in some way?



Answer



Let's call the OP's diagram Σ(xy), which can be rewritten as Σ(xy)=Adpxdpyδ(pxpy)ei(pxxpyy)DF(px)DF(py),

where A is a constant which comes from the two loops contained in the diagram, and DF(p)=(p2m2+iϵ)1.


Playing the game of the second part of the question (i.e. replacing DF(z1y) by DF(yz1)), we would end up with Σ(xy)=Adpxdpyδ(px+py)ei(pxx+pyy)DF(px)DF(py),

which is obviously equivalent to the first expression, by the change of variable pypy. There is therefore no contradiction, and both way of doing the calculation are equivalent.


What confuses the OP is the question of momentum conservation, which looks strange in the second equation. This is because we should look at momentum conservation in momentum space (whereas we are still in real space here), i.e. we should compute Σ(p1,p2)=dxdyeip1xip2yΣ(xy).


Starting from either expression, we obtain Σ(p1,p2)=δ(p1p2)ADF(p1)DF(p2),

which obviously conserve momentum. One usually factors out the delta function and introduce the function Σ(p)=ADF(p)DF(p) such that Σ(px,py)=δ(pxpy)Σ(px).


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